Question

The value of ${{\sin }^{-1}}\left( \cos \left( {{\cos }^{-1}}\cos x+{{\sin }^{-1}}\sin x \right) \right)$, where $x\in \left( \dfrac{\pi }{2},\pi \right)$ is equal to
[a] $\dfrac{\pi }{2}$
[b] $\pi $
[c] $-\pi $
[d] $-\dfrac{\pi }{2}$

Answer 1

Hint: Use the fact that if $\sin x=\sin y,$ then $x=n\pi +{{\left( -1 \right)}^{n}}y,n\in \mathbb{Z}$ and ${{\sin }^{-1}}x\in \left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right]$ and if $\cos x=\cos y$, then $x=2n\pi \pm y,n\in \mathbb{Z}$ and ${{\cos }^{-1}}x\in \left[ 0,\pi \right]$. Assume $u={{\cos }^{-1}}\left( \cos x \right)$ and hence prove that $u=2n\pi \pm x,n\in \mathbb{Z}$. Find the suitable value of n such that $u\in \left[ 0,2\pi \right]$. Hence find the value of u. Similarly, assume $v={{\sin }^{-1}}\left( \sin x \right)$ and hence prove that $v=n\pi +{{\left( -1 \right)}^{n}}x,n\in \mathbb{Z}$. Find the suitable value of n such that $v\in \left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right]$. Hence find the value of v. Hence find the value of $\cos \left( {{\cos }^{-1}}\cos x+{{\sin }^{-1}}\sin x \right)$ and hence evaluate the given expression.

Complete step-by-step answer:
Let $u={{\cos }^{-1}}\cos x$
We know that if $y={{\cos }^{-1}}x\Rightarrow x=\cos y$
Hence, we have
$\cos u=\cos x$
We know that if $\cos x=\cos y$, then $x=2n\pi \pm y,n\in \mathbb{Z}$
Hence, we have
$u=2n\pi \pm x,n\in \mathbb{Z}$
We know that ${{\cos }^{-1}}x\in \left[ 0,\pi \right]$
Hence, we have $u\in \left[ 0,\pi \right]$
Now since $x\in \left( \dfrac{\pi }{2},\pi \right)$, we have \[x\in \left[ 0,\pi \right]\]
Hence, we have
$u=x$
Now, let $v={{\sin }^{-1}}\sin x$
Hence, we have
$\sin v=\sin x$
We know that if $\sin x=\sin y,$ then $x=n\pi +{{\left( -1 \right)}^{n}}y,n\in \mathbb{Z}$
Hence, we have
$v=n\pi +{{\left( -1 \right)}^{n}}x$
Since ${{\sin }^{-1}}x\in \left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right]$, we have
$v\in \left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right]$
Now since $x\in \left( \dfrac{\pi }{2},\pi \right),$ we have $\pi -x\in \left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right]$
Hence, we have
$v=\pi -x$
Hence, we have
${{\sin }^{-1}}\sin x+{{\cos }^{-1}}\cos x=v+u=\pi -x+x=\pi $
Hence, we have
$\cos \left( {{\sin }^{-1}}\sin x+{{\cos }^{-1}}\cos x \right)=\cos \pi =-1$
We know that ${{\sin }^{-1}}\left( -x \right)=-{{\sin }^{-1}}x$
Hence, we have
$\sin \left( \cos \left( {{\sin }^{-1}}\sin x+{{\cos }^{-1}}\cos x \right) \right)={{\sin }^{-1}}\left( -1 \right)=-{{\sin }^{-1}}1$
We know that ${{\sin }^{-1}}\left( 1 \right)=\dfrac{\pi }{2}$
Hence, we have
$\sin \left( \cos \left( {{\sin }^{-1}}\sin x+{{\cos }^{-1}}\cos x \right) \right)=-\dfrac{\pi }{2}$
Hence option [d] is correct.

Note: Alternative solution:
We know that
${{\sin }^{-1}}\sin \left( x \right)=\left\{ \begin{matrix}
   \vdots \\
   -\pi -x,x\in \left[ \dfrac{-3\pi }{2},\dfrac{-\pi }{2} \right] \\
   x,x\in \left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right] \\
   \pi -x,x\in \left[ \dfrac{\pi }{2},\dfrac{3\pi }{2} \right] \\
   \vdots \\
\end{matrix} \right.$ and ${{\cos }^{-1}}\cos x=\left\{ \begin{matrix}
   \vdots \\
   2\pi +x,x\in \left[ -2\pi ,-\pi \right] \\
   x,x\in \left[ 0,\pi \right] \\
   2\pi -x,x\in \left[ \pi ,2\pi \right] \\
   \vdots \\
\end{matrix} \right.$
Using these definitions, we can find the value of the above expression.
Graph of ${{\sin }^{-1}}\sin x$:

Graph of ${{\cos }^{-1}}\cos x$: